What is the derivative of #f(t) = (-1/(t+1) , e^(2t-1) ) #?

2 Answers
Shwetank Mauria
May 7, 2017

#(dy)/(dx)=(t+1)^2e^(2t-1)#

Explanation:

This is a parametric form of equation i.e. #f(t)=(x(t),y(t))#

In such a case #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))#

Here #x(t)=-1/(t+1)# and #(dx)/(dt)=-(-1)/(t+1)^2=1/(t+1)^2#

and #y(t)=e^(2t-1)# and #(dy)/(dt)=2e^(2t-1)#

Hence #(dy)/(dx)=(2e^(2t-1))/(1/(t+1)^2)=(t+1)^2e^(2t-1)#

mason m
May 7, 2017

For a parametric function #f(t)=(x(t),y(t))#, the function's derivative is just given by the #x# and #y# components' derivatives: #f'(t)=(x'(t),y'(t))#.

So:

#f(t)=(-(t+1)^-1,e^(2t-1))#

#f'(t)=(-(-(t+1)^-2),2e^(2t-1))#

#f'(t)=(1/(t+1)^2,2e^(2t-1))#

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